Problem: Simplify; express your answer in exponential form. Assume $q\neq 0, r\neq 0$. $\dfrac{{(q^{-5})^{3}}}{{(q^{-2}r^{-3})^{5}}}$
Answer: To start, try working on the numerator and the denominator independently. In the numerator, we have ${q^{-5}}$ to the exponent ${3}$ . Now ${-5 \times 3 = -15}$ , so ${(q^{-5})^{3} = q^{-15}}$ In the denominator, we can use the distributive property of exponents. ${(q^{-2}r^{-3})^{5} = (q^{-2})^{5}(r^{-3})^{5}}$ Simplify using the same method from the numerator and put the entire equation together. $\dfrac{{(q^{-5})^{3}}}{{(q^{-2}r^{-3})^{5}}} = \dfrac{{q^{-15}}}{{q^{-10}r^{-15}}}$ Break up the equation by variable and simplify. $\dfrac{{q^{-15}}}{{q^{-10}r^{-15}}} = \dfrac{{q^{-15}}}{{q^{-10}}} \cdot \dfrac{{1}}{{r^{-15}}} = q^{{-15} - {(-10)}} \cdot r^{- {(-15)}} = q^{-5}r^{15}$.